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How to calculate the following limit - I'm stuck!

  1. Jan 4, 2009 #1
    1. The problem statement, all variables and given/known data

    what is the limit of factoring.jpg as x approaches 9?

    2. Relevant equations



    3. The attempt at a solution

    factoringsol.jpg I'm stuck please help...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 4, 2009 #2
    Re: factoring

    I would do it this way:
    remember that:
    a[tex]^{2}[/tex]-b[tex]^{2}[/tex]=(a+b)(a-b)
    so
    a-b=[tex]\frac{(a^2-b^2}{(a+b)}[/tex]

    so take the numerator and denominator of your problem and treat each as a-b
    so:
    [tex]\sqrt{x}-3[/tex] = [tex]\frac{x-9}{\sqrt{x}+3}[/tex] (1)

    and
    [tex]\sqrt{1+\sqrt{x}}-2[/tex] = [tex]\frac{\sqrt{x} - 3}{\sqrt{1+\sqrt{x}}+2}[/tex] (2)

    so your limit is now (1)/(2): which gives:

    [tex]\frac{x-9}{\sqrt{x}+3}\times\frac{\sqrt{1+\sqrt{x}}+2}{\sqrt{x}-3}[/tex]

    Is this clear?
    Do you know how to continue from here?
     
  4. Jan 4, 2009 #3
    Re: factoring

    continuation...
    sol2.jpg

    thus, the limit of giv.jpg as x approaches 9 is 4. Is this correct?
     
  5. Jan 4, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: factoring

    Correct.
     
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